Problem 21 · 2013 AMC 8
Hard
Counting & Probability
lattice-pathsmultiplication-principle
Samantha lives 2 blocks west and 1 block south of the southwest corner of City Park. Her school is 2 blocks east and 2 blocks north of the northeast corner of City Park. On school days she bikes on streets to the southwest corner of City Park, then takes a diagonal path through the park to the northeast corner, and then bikes on streets to school. If her route is as short as possible, how many different routes can she take?
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Answer: E — 18 routes.
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Hint 1 of 2
The forced diagonal through the park splits the trip into three separate legs that don't interfere with each other. Count each leg's shortest routes on its own, then combine — how do independent choices multiply?
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Hint 2 of 2
A shortest grid route is just an arrangement of fixed moves (so many easts, so many norths). Number of routes = C(total steps, steps in one direction). And independent stages multiply.
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Approach: count each leg, then multiply (independent stages)
- Leg 1, home → SW corner: 2 easts and 1 north in some order. Choosing where the single north goes among 3 steps gives C(3, 1) = 3 routes.
- Leg 2, the diagonal through the park: a single forced path = 1 way (this is what cleanly separates the two grids).
- Leg 3, NE corner → school: 2 easts and 2 norths, C(4, 2) = 6 routes.
- The legs are independent, so multiply: 3 × 1 × 6 = 18.
- You'll see this again: shortest-path counts on a grid are always "choose which steps are north," and independent stages of a journey multiply — the multiplication principle.
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